Tag Archives: fields

An algebraic field extension \(K \subset L\) is said to be normal if every irreducible polynomial, either has no root in \(L\) or splits into linear factors in \(L\).

One can prove that if \(L\) is a normal extension of \(K\) and if \(E\) is an intermediate extension (i.e., \(K \subset E \subset L\)), then \(L\) is a normal extension of \(E\).

However a normal extension of a normal extension may not be normal and the extensions \(\mathbb Q \subset \mathbb Q(\sqrt{2}) \subset \mathbb Q(\sqrt[4]{2})\) provide a counterexample. Let’s prove it.

As a short lemma, we prove that a quadratic extension \(k \subset K\) , i.e. an extension of degree two is normal. Suppose that \(P\) is an irreducible polynomial of \(k[x]\) with a root \(a \in K\). If \(a \in k\) then the degree of \(P\) is equal to \(1\) and we’re done. Otherwise \((1, a)\) is a basis of \(K\) over \(k\) and there exist \(\lambda, \mu \in k\) such that \(a^2 = \lambda a +\mu\). As \(a \notin k\), \(Q(x)= x^2 – \lambda x -\mu\) is the minimal polynomial of \(a\) over \(k\). As \(P\) is supposed to be irreducible, we get \(Q = P\). And we can conclude as \[
Q(x) = (x-a)(x- \lambda +a).\]

The entensions \(\mathbb Q \subset \mathbb Q(\sqrt{2})\) and \(\mathbb Q(\sqrt{2}) \subset \mathbb Q(\sqrt[4]{2})\) are quadratic, hence normal according to previous lemma and \(\sqrt[4]{2}\) is a root of the polynomial \(P(x)= x^4-2\) of \(\mathbb Q[x]\). According to Eisenstein’s criterion \(P\) is irreducible over \(\mathbb Q\). However \(\mathbb Q(\sqrt[4]{2}) \subset \mathbb R\) while the roots of \(P\) are \(\pm \sqrt[4]{2}, \pm i \sqrt[4]{2}\) and therefore not all real. We can conclude that \(\mathbb Q \subset \mathbb Q(\sqrt[4]{2})\) is not normal.

Let’s describe an example of a field \(K\) which is of degree \(2\) over two distinct subfields \(M\) and \(N\) respectively, but not algebraic over \(M \cap N\).

Let \(K=F(x)\) be the rational function field over a field \(F\) of characteristic \(0\), \(M=F(x^2)\) and \(N=F(x^2+x)\). I claim that those fields provide the example we’re looking for.

\(K\) is of degree \(2\) over \(M\) and \(N\)

The polynomial \(\mu_M(t)=t^2-x^2\) belongs to \(M[t]\) and \(x \in K\) is a root of \(\mu_M\). Also, \(\mu_M\) is irreducible over \(M=F(x^2)\). If that wasn’t the case, \(\mu_M\) would have a root in \(F(x^2)\) and there would exist two polynomials \(p,q \in F[t]\) such that \[
p^2(x^2) = x^2 q^2(x^2)\] which cannot be, as can be seen considering the degrees of the polynomials of left and right hand sides. This proves that \([K:M]=2\). Considering the polynomial \(\mu_N(t)=t^2-t-(x^2+x)\), one can prove that we also have \([K:N]=2\).

We have \(M \cap N=F\)

The mapping \(\sigma_M : x \mapsto -x\) extends uniquely to an \(F\)-automorphism of \(K\) and the elements of \(M\) are fixed under \(\sigma_M\). Similarly, the mapping \(\sigma_N : x \mapsto -x-1\) extends uniquely to an \(F\)-automorphism of \(K\) and the elements of \(N\) are fixed under \(\sigma_N\). Also \[
(\sigma_N\circ\sigma_M)(x)=\sigma_N(\sigma_M(x))=\sigma_N(-x)=-(-x-1)=x+1.\] An element \(z=p(x)/q(x) \in M \cap N\) where \(p(x),q(x)\) are coprime polynomials of \(K=F(x)\) is fixed under \(\sigma_M \circ \sigma_N\). Therefore following equality holds \[
\frac{p(x)}{q(x)}=z=(\sigma_2\circ\sigma_1)(z)=\frac{p(x+1)}{q(x+1)},\] which is equivalent to \[
p(x)q(x+1)=p(x+1)q(x).\] By induction, we get for \(n \in \mathbb Z\) \[
p(x)q(x+n)=p(x+n)q(x).\] Assume \(p(x)\) is not a constant polynomial. Then it has a root \(\alpha\) in some finite extension \(E\) of \(F\). As \(p(x),q(x)\) are coprime polynomials, \(q(\alpha) \neq 0\). Consequently \(p(\alpha+n)=0\) for all \(n \in \mathbb Z\) and the elements \(\alpha +n\) are all distinct as the characteristic of \(F\) is supposed to be non zero. This implies that \(p(x)\) is the zero polynomial, in contradiction with our assumption. Therefore \(p(x)\) is a constant polynomial and \(q(x)\) also according to a similar proof. Hence \(z\) is constant as was supposed to be proven.

Finally, \(K=F(x)\) is not algebraic over \(F=M \cap N\) as \((1,x, x^2, \dots, x^n, \dots)\) is independent over the field \(F\) which concludes our claims on \(K, M\) and \(N\).

Consider a vector space \(V\) over a field \(F\). A subspace \(W \subseteq V\) is an additive subgroup of \((V,+)\). The converse might not be true.

If the characteristic of the field is zero, then a subgroup \(W\) of \(V\) might not be an additive subgroup. For example \(\mathbb R\) is a vector space over \(\mathbb R\) itself. \(\mathbb Q\) is an additive subgroup of \(\mathbb R\). However \(\sqrt{2}= \sqrt{2}.1 \notin \mathbb Q\) proving that \(\mathbb Q\) is not a subspace of \(\mathbb R\).

Another example is \(\mathbb Q\) which is a vector space over itself. \(\mathbb Z\) is an additive subgroup of \(\mathbb Q\), which is not a subspace as \(\frac{1}{2} \notin \mathbb Z\).

Finally an additive subgroup of a vector space over any finite field is not always a subspace. For a counterexample, take the non-prime finite field \(\mathbb F_{p^2}\) (also named \(\text{GF}(p^2)\)). \(\mathbb F_{p^2}\) is also a vector space over itself. The prime finite field \(\mathbb F_p \subset \mathbb F_{p^2}\) is an additive subgroup that is not a subspace of \(\mathbb F_{p^2}\).

Let’s recall that an ordered field \(K\) is said to be Archimedean if for any \(x \in K\) there exists a natural number \(n\) such that \(n > x\).

The ordered fields \(\mathbb Q\) or \(\mathbb R\) are Archimedean. We introduce here the example of an ordered field which is not Archimedean. Let’s consider the field of rational functions
\[\mathbb R(x) = \left\{\frac{S(x)}{T(x)} \ | \ S, T \in \mathbb R[x] \right\}\] For \(f(x)=\frac{S(x)}{T(x)} \in \mathbb R(x)\) we can suppose that the polynomials have a constant polynomial greatest common divisor.

Now we define \(P\) as the set of elements \(f(x)=\frac{S(x)}{T(x)} \in \mathbb R(x)\) in which the leading coefficients of \(S\) and \(T\) have the same sign.

One can verify that the subset \(P \subset \mathbb R(x)\) satisfies following two conditions:

ORD 1

Given \(f(x) \in \mathbb R(x)\), we have either \(f(x) \in P\), or \(f(x)=0\), or \(-f(x) \in P\), and these three possibilities are mutually exclusive. In other words, \(\mathbb R(x)\) is the disjoint union of \(P\), \(\{0\}\) and \(-P\).

ORD 2

For \(f(x),g(x) \in P\), \(f(x)+g(x)\) and \(f(x)g(x)\) belong to \(P\).

This means that \(P\) is a positive cone of \(\mathbb R(x)\). Hence, \(\mathbb R(x)\) is ordered by the relation
\[f(x) > 0 \Leftrightarrow f(x) \in P.\]

Now let’s consider the rational fraction \(h(x)=\frac{1}{x} \in \mathbb R(x)\). \(h(x)\) is a positive element, i.e. belongs to \(P\). And for any \(n \in \mathbb N\), we have
\[h(x)-n=\frac{1}{x}-n=\frac{x-n}{x} \in P\] as the leading coefficients of \(x\) and \(x-n\) are both equal to \(1\). Therefore, we have \(h(x) > n\) for all \(n \in \mathbb N\), proving that \(\mathbb R(x)\) is not Archimedean.

We also know rings or fields like integers modulo \(n\) (with \(n \ge 2\)) \(\mathbb Z_n\) or the finite field \(\mathbb F_q\) with \(q=p^r\) elements where \(p\) is a prime.
We provide below examples of infinite rings or fields with positive characteristic.

Infinite rings with positive characteristic

Consider the ring \(\mathbb Z_n[X]\) of polynomials in one variable \(X\) with coefficients in \(\mathbb Z_n\) for \(n \ge 2\) integer. It is an infinite ring since \(\mathbb X^m \in \mathbb{Z}_n[X]\) for all positive integers \(m\), and \(X^r \neq X^s\) for \(r \neq s\). But the characteristic of \(\mathbb Z_n[X]\) is clearly \(n\).

Another example is based on product of rings. If \(I\) is an index set and \((R_i)_{i \in I}\) a family of rings, one can define the product ring \(\displaystyle \prod_{i \in I} R_i\). The operations are defined the natural way with \((a_i)_{i \in I} + (b_i)_{i \in I} = (a_i+b_i)_{i \in I}\) and \((a_i)_{i \in I} \cdot (b_i)_{i \in I} = (a_i \cdot b_i)_{i \in I}\). Fixing \(n \ge 2\) integer and taking \(I = \mathbb N\), \(R_i = \mathbb Z_n\) for all \(i \in I\) we get the ring \(\displaystyle R = \prod_{k \in \mathbb N} \mathbb Z_n\). \(R\) multiplicative identity is the sequence with all terms equal to \(1\). The characteristic of \(R\) is \(n\) and \(R\) is obviously infinite. Continue reading Infinite rings and fields with positive characteristic→

We give here an example of a division ring which is not commutative. According to Wedderburn theorem every finite division ring is commutative. So we must turn to infinite division rings to find a non-commutative one, i.e. a skew field.

We raise here the following question: “can a vector space \(E\) be written as a finite union of proper subspaces”?

Let’s consider the simplest case, i.e. writing \(E= V_1 \cup V_2\) as a union of two proper subspaces. By hypothesis, one can find two non-zero vectors \(v_1,v_2\) belonging respectively to \(V_1 \setminus V_2\) and \(V_2 \setminus V_1\). The relation \(v_1+v_2 \in V_1\) leads to the contradiction \(v_2 = (v_1+v_2)-v_1 \in V_1\) while supposing \(v_1+v_2 \in V_2\) leads to the contradiction \(v_1 = (v_1+v_2)-v_2 \in V_2\). Therefore, a vector space can never be written as a union of two proper subspaces.